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Chapter 5: Problem 2

If \(f\) is an even function, why is \(\int_{-a}^{a} f(x) d x=2 \int_{0}^{a} f(x) d x ?\)

Short Answer

Expert verified
Question: Show that the integral of an even function over a symmetric interval [-a, a] is equal to twice the integral over [0, a]. Answer: If f is an even function, then \(\int_{-a}^{a} f(x) dx = 2\int_{0}^{a} f(x) dx\).

Step by step solution

01

Define an even function

An even function is a function that satisfies the property \(f(x) = f(-x)\) for all x in its domain.
02

Rewrite the given integral using the substitution method

We will use the substitution method to rewrite the integral. Let \(u=-x\). Then, \(du=-dx\). Since the integral is from \(-a\) to \(a\), we also have to change the limits of integration. For \(x=-a\), we have \(u=a\) and for \(x=a\), we have \(u=-a\). Rewriting the integral, we get: $$\int_{-a}^{a} f(x) dx=-\int_{a}^{-a} f(-u) du$$
03

Change the limits of integration

Now we need to reverse the limits of integration, which means multiplying by -1 and switching the limits: $$-\int_{a}^{-a} f(-u) du=\int_{-a}^{a} f(-u) du$$
04

Use the definition of an even function

Since \(f\) is an even function, we know that \(f(-u)=f(u)\). Substituting this into the integral, we get: $$\int_{-a}^{a} f(-u) du=\int_{-a}^{a} f(u) du$$
05

Split the integral into two parts

Now, we need to split the integral from -a to a into two parts: one from -a to 0 and one from 0 to a: $$\int_{-a}^{a} f(u) du=\int_{-a}^{0} f(u) du + \int_{0}^{a} f(u) du$$
06

Use the result from Step 3

Recall that in Step 3, we found \(\int_{-a}^{a} f(x) dx=\int_{-a}^{a} f(-u) du\). Using this result, we can write: $$\int_{-a}^{a} f(x) dx=\int_{-a}^{0} f(u) du + \int_{0}^{a} f(u) du$$
07

Combine the integrals into a single integral

Since both integrals have the same integration limits, we can combine them into a single integral: $$\int_{-a}^{a} f(x) dx = 2\int_{0}^{a} f(u) du$$
08

Change back to the original variable

Finally, we change the variable back from \(u\) to \(x\): $$\int_{-a}^{a} f(x) dx = 2\int_{0}^{a} f(x) dx$$ Thus, we have shown that if \(f\) is an even function, then \(\int_{-a}^{a} f(x) dx = 2\int_{0}^{a} f(x) dx\).

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